100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________

100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M F

1.	100 mL of 0.01 M XO_(4)^(-) is reduced to X^(n+) by 100 mL of 0.05 M Fe^(2+) in acidic medium. Thusoxidation state of X in X^(n+) is ______________
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`100xx0.05xx1=100xx0.01xx(7-n)n=2`</body></html>